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Let $k\subset F\subseteq k(x_{1},x_{2},...,x_{n})$ where $k$ and $F$ are the fields and $x_{1},x_{2},...,x_{n}$ are transcendental over $k$. Can we express $F$ in terms of or function of $x_{1},x_{2},...,x_{n}$ ?. Thanks

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up vote 6 down vote accepted

The field extension $k(x_1,...,x_n)/k$ is finitely generated, hence so is $F/k$ (see $\S 11.4$ of these notes). That is, there are finitely many rational functions $f_1(x_1,\ldots,x_n),\ldots,f_m(x_1,\ldots,x_n)$ such that $F = k(f_1,\ldots,f_m)$.

When $n = 1$ it is further the case that we may always take $m = 1$, and then $F = k(f)$ is again a rational function field: Luroth's Theorem. For $n > 1$ a subfield of a rational function field need not be rational, and I can't think of anything nice to say about the extension $F/k$ except that it is finitely generated -- at least not without bringing in some fairly sophisticated algebraic geometry.

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@ Pete, Thank you very much. I got it. – Rajesh Feb 20 '13 at 4:13

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